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A253596 Numbers k such that A002313(m) is the greatest prime divisor of k^2 + 1 and A002313(m+1) is the greatest prime divisor of (k+1)^2 + 1 for some m.

%I A253596 #11 Aug 05 2021 20:40:16
%S A253596 1,7,31,293,1936,2244,4158,5744,11573,25242,285202,339354
%N A253596 Numbers k such that A002313(m) is the greatest prime divisor of k^2 + 1 and A002313(m+1) is the greatest prime divisor of (k+1)^2 + 1 for some m.
%C A253596 A002313 contains the primes congruent to 1 or 2 (mod 4).
%C A253596 The corresponding indices m in A002313 are 1, 2, 6, 13, 69, 65, 322, 199, 130, 46, 1471, 866, ...
%C A253596 The corresponding primes A002313(m) are 2, 5, 37, 101, 809, 761, 4877, 2777, 1709, 509, 26821, 14957, ...
%e A253596 31 is in the sequence because 31^2 + 1 = 2*13*37 and 32^2 + 1 = 5*5*41 with the property that 37 = A002313(6) and 41 = A002313(7).
%p A253596 with(numtheory): nn:=500000:print(1):
%p A253596 for n from 1 to nn do:
%p A253596    p:=n^2+1:x:=factorset(p):n0:=nops(x):p1:=x[n0]:
%p A253596    q:=(n+1)^2+1:y:=factorset(q):n1:=nops(y):p2:=y[n1]:ii:=0:
%p A253596      for j from 2 by 2 to 1000 while(ii=0) do:
%p A253596       pp:=p1+j:
%p A253596       if type(pp,prime)=true and irem(pp,4)=1
%p A253596       then
%p A253596       p3:=pp:ii:=1:
%p A253596       else
%p A253596       fi:
%p A253596     od:
%p A253596     if p3=p2
%p A253596     then
%p A253596     print(n):
%p A253596      else
%p A253596      fi:
%p A253596     od:
%t A253596 lst={};Do[If[Mod[Prime[i],4]==1||Mod[Prime[i],4]==2,AppendTo[lst,Prime[i]]],{i,1,1000}];Do[Do[If[FactorInteger[n^2+1][[-1]][[1]]==Part[lst,j]&&FactorInteger[(n+1)^2+1][[-1]][[1]]==Part[lst,j+1],Print[n]],{n,1,20000}],{j,1,999}]
%Y A253596 Cf. A002313, A014442.
%K A253596 nonn,more
%O A253596 1,2
%A A253596 _Michel Lagneau_, Jan 05 2015